If r2c8 = 6 then r2c5<>6, r7c5 must be 6 making r7c4=7
If r7c4=6 then r7c5<>6 and r2c5 must be 6 making r2c8=7

Either way the 67s act as a pair, so r2c4<>7 or 6 and so =5

Have I made an error?

Hugh

I don't know if you've made an error or not because a glaring weakness of mine is the inability to follow along with that style of reasoning. However, your premise of either r2c8 or r7c4 being 6 puzzles me because I don't see anything that says either of those two cells must be 6. Perhaps someone else will come along and straighten me out.

But my original response was about the statement about the W-Wing, "Didn't even need to look for the 67 W-Wing
in c5 setting r2c4 to 5!"

The W-Wing consists of 67 cells in r2c8 and r7c4, connected by the strong link on 6 in c5. That says at least one of r2c8 and r7c4 must be 7, thus making r2c4<>7.

In other words, buddies of a and b cannot be 6, of f and a cannot be 7, etc. I have marked the eliminations I can see.

Some M-wings are also often cycles.

Here is another XY cycle in the same grid, found by Sudoku Susser:

Code:

Found a 7-link Simple Forcing Loop. If we assume that square [x=2,y=9] is <3> then we can make the following chain of conclusions:

[x=5,y=9] must be <7>, which means that
[x=4,y=7] must be <6>, which means that
[x=4,y=3] must be <4>, which means that
[x=7,y=3] must be <6>, which means that
[x=7,y=8] must be <3>, which means that
[x=2,y=8] must be <6>, which means that
[x=2,y=9] must be <3>.

This forms a inherently bi-directional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:

One of [x=2,y=9] and [x=5,y=9] must be <3>.
One of [x=5,y=9] and [x=4,y=7] must be <7>.
One of [x=4,y=7] and [x=4,y=3] must be <6>.
One of [x=4,y=3] and [x=7,y=3] must be <4>.
One of [x=7,y=3] and [x=7,y=8] must be <6>.
One of [x=7,y=8] and [x=2,y=8] must be <3>.
One of [x=2,y=8] and [x=2,y=9] must be <6>.

Thus we can deduce that:

[x=8,y=9] - cannot contain <3> because of [x=2,y=9] and [x=5,y=9].
[x=5,y=7] - cannot contain <7> because of [x=5,y=9] and [x=4,y=7].
[x=4,y=2] - cannot contain <6> because of [x=4,y=7] and [x=4,y=3].

I don't know if you've made an error or not because a glaring
weakness of mine is the inability to follow along with that style of reasoning.
However, your premise of either r2c8 or r7c4 being 6 puzzles me because I
don't see anything that says either of those two cells must be 6. Perhaps
someone else will come along and straighten me out.

Yes Marty- you've seen the flaw in my logic- of course (without the
strong link in C9 on 7 from later contributors) there is no reason
why one of these squares must be 6. I might have got the right
answer, but through wrong reasoning!